DISTILLATION - جامعة نزوى · 24/02/2015 3 VAPOR – LIQUID ... Flash distillation is done by vaporizing a definite fraction of a liquid mixture in such a way that the evolved

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24/02/2015

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DISTILLATION

Definition & General Description of the process

Vapor – liquid equilibrium relationship Flash, Batch and Fractionating Distillations

Physical concept of Distillation

Definition & General Description of the process

Separating the various components of a liquid solution

Depends upon the distribution of these components between a vapor phase and a liquid phase.

All components presents in both phases.

Distillation is done by vaporizing a

definite fraction of a liquid mixture in such a way that the evolved vapor is in equilibrium with the residual liquid.

The equilibrium vapor is then separated from the equilibrium residual liquid by condensing the vapor.

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Carried out by either two principal method

First method: based on the production of a vapor by boiling the liquid mixture to be separated and condensing the vapors without allowing any liquid to return to the still – no reflux (Eg. Flash distillation)

Second method: Based on the return part of the condensate to the still under such condition that this returning liquid is brought into intimate contact with the vapors on their way to the condenser – conducted as continuous or batch process (Eg. Continuous distillation).

Physical concept of Distillation

VAPOR – LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID MIXTURE

VAPOR – LIQUID EQUILIBRIUM

DEFINITION

EVAPORATION – The phase transformation processes from liquid to gas or vapor phase.

VOLATILITY – The tendency of liquid to change form to gas

PREDICTION OF VAPOR – LIQUID EQUILIBRIUM COMPOSITIONS FOR ORDINARY BINARY MIXTURES

RELATIVE VOLATILITY OF A MIXTURE

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VAPOR – LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID MIXTURE

A binary liquid mixture – consist of two different liquids. Can be classified into homogeneous mixtures or non – homogeneous (heterogeneous) mixtures.

Low boiler liquid (A) – liquid that vaporized easily (low boiling point or high vapor pressure)

High boiler liquid (B) – liquid which have higher boiling point or low vapor pressure

Homogeneous mixtures – mix at all proportions resulting in one continuous phase.

Heterogeneous mixtures – do not mix uniformly resulting in more than one distinct phases.

Equilibrium curve – shows the relationship between composition of residual liquid and vapor that are in dynamic phase equilibrium. This curve will be very useful in calculations to predict the number of the ideal stages required for a specified distillation process.

Equilibrium mole fraction of A in vapor is larger than mole fraction of A in liquid phase. This is expected since that A has lower boiling point than B, A would vaporize more than B.

Graphical mass balances for equilibrium systems

Determination the amounts of liquid and vapor when two

phase mixture separates (Lever arm rule)

ML

MN

xm

my

V

L

VLF

VyLxFm

LA

AN

NLA

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Raoult’s Law for ideal solution & Dalton’s Law of partial pressure can be manipulated in order to calculate compositions of liquid and vapor, which are in equilibrium.

Raoult’s Law – the partial pressure of a component in the vapor phase is equal to the mole fraction of the component in the liquid multiplied by its pure vapor pressure at that temperature. = partial pressure of A in vapor phase = mole fraction of A in liquid phase

= vapor pressure of A at that temperature

Ap

Ax

o

AP

o

AAA Pxp


For a mixture of the different gases inside a close container, Dalton’s law stated that the resultant total pressure of the container is the summation of partial pressures of each of all the gases that make up the gas mixture.

Dalton also stated that the partial pressure of gas (pA) is:

= partial pressure of A in vapor phase

= mole fraction of A in vapor phase

= Total pressure of the system

BAT ppP

Ay

TP

Ap

TAA Pyp

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For a mixture of the different gases inside a close container, Dalton’s law stated that the resultant total pressure of the container is the summation of partial pressures of each of all the gases that make up the gas mixture.

Dalton also stated that the partial pressure of gas (pA) is:

= partial pressure of A in vapor phase

= mole fraction of A in vapor phase

= Total pressure of the system

BAT ppP

Ay

TP

Ap

TAA Pyp

Separations of components by distillation process depend on the differences in volatilities of components that make up the solution to be distilled.

The greater difference in their volatility, the better is separation by heating (distillation). Conversely if their volatility differ only slightly, the separation by heating becomes difficult.


Equilibrium diagram for n-heptane n-octane mixture at 101.3 kPa

y = x

0.00

0.20

0.40

0.60

0.80

1.00

0.00 0.20 0.40 0.60 0.80 1.00

xA

yA

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In figure above, the greater the distance between the equilibrium line and the 45o line, the greater the difference between the vapor composition and a liquid composition. Separation is more easily made.

A numerical measure of ‘how easy’ separation – relative volatility, AB. Relative volatility, AB - ratio of the concentration of A in the vapor to the

concentration of A in the liquid divided by the ratio of the concentration B in the vapor to the concentration of B in the liquid.

AB - relative volatility of A with respect to B in the binary system.

If the system obeys Raoult’s Law for an ideal system: and ,


AA

AA

BB

AAAB

xy

xy

xy

xy

11

T

AA

AP

xPy

T

BB

BP

xPy

B

A

ABP

P A

A

Ax

xy

11

For non – ideal solution, the values α of change with temperature while for ideal solution, the values of doesn’t change with temperature. For a solution that approaches ideal solution, its would fairly constant.

A

AA

y

yx

1

Flash distillation – a single stage process because it has only one vaporization stage (means one liquid phase is exposed to one vapor phase.

Flash distillation is done by vaporizing a definite fraction of a liquid mixture in such a way that the evolved vapor is in equilibrium with the residual liquid.

The equilibrium vapor is then separated from the equilibrium residual liquid by condensing the vapor.

Flash distillation can be either by batch or continuous

FLASH (EQUILIBRIUM) DISTILLATION

Figure 1: A typical flow chart of a continuous flash distillation

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As illustrated in Figure 1, a liquid mixture feed, with initial mole fraction of A at xF, is pre-heated by a heater and its pressure is then reduced by an expansion valve.

At lower pressure, the hot liquid will boil at lower temperature inside a separator drum. In contrast to that of a simple batch distillation, the evolved vapor is allowed to attain equilibrium with remaining liquid inside the flash drum (separator).

Now we are interested to predict the composition (x* and y*) of these vapor and liquid that are in equilibrium with each other.


Prediction of an equilibrium composition of vapor and liquid for a flash distillation of an ordinary binary mixture

Consider a binary liquid mixture which is to be separated by a flash distillation process as shown in Figure 1. Binary is composed of two components; A as low boiler (more volatile) and B as a high boiler.

Total mass balance:

Mass balance on more volatile component (A):

* denote ‘at state of equilibrium Eq. (1)

Assign f to denote fraction of the feed which has been vaporized. f is defined

mathematically as:


LVF

LxVyFxF **

F

Lx

F

VyxF **

F

Vf

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And substituting f in the total mass balance:

Eq. (2)

Substituting Eq. 2 into Eq. (1) to obtain: Eq. (3)

Eq. (3) called operating line equation (y = mx + c) for a flash distillation process.


LVF F

L

F

V1

F

Lf1 f1

F

L

)(** f1xfyxF

f

x

f

x1fy F

**

Since xF and f do not change throughout the process, only y* and x* are the variables. A straight line with a negative slope should be obtained from the operating line. The line gives all locus point (x*, y*) that satisfy the operating line.


Figure 2: Graphical method to determine equilibrium compositions in a flash

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A liquid mixture containing 70 mol% n-heptane (A) and 30 mol% n-octane at 30OC is to be continuously flash – vaporized at the standard atmospheric pressure to vaporize 60 mol% of the feed.

What will be the compositions of the vapor and liquid and the temperature in

the separator for an equilibrium stage? The equilibrium data for n-heptane-n-octane mixture at 1 atm and 30OC is given

as follows:


Example 1

T (K) xA yA

371.6 1.000 1.000

374.0 0.825 0.920

377.0 0.647 0.784

380.0 0.504 0.669

383.0 0.387 0.558

386.0 0.288 0.449

389.0 0.204 0.342

392.0 0.132 0.236

395.0 0.068 0.132

398.2 0.000 0.000


Solution

C378 mequilibriuat separator theof re temperatu the

0.76y* in vapor, heptane-n offraction mol mequilibriu

0.62 x*liquid,in heptane-n offraction mol mequilibriu

graph, in theshown

as curve mequilibriu theand line operating theofon intersecti theFrom

167.1*667.0*

6.0

7.0

6.0

*16.0* :line operating The

* xand *y liquid; and vapor of nscompositio mequilibriu find want toWe

0.610060FVf moles 601006.0

7.0 xGiven,

(F) feed liquid of moles 100 Basis

O

F

xy

xy

V

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Determination of vapor-liquid equilibrium composition

for a flash distillation of n-heptane-n-octane mixture

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x,mol fraction n-heptane in liquid

y,

mo

l fr

acti

on

of

n-h

ep

tan

e i

n v

ap

or

Simple batch distillation which is also known as differential distillation refer to a batch distillation in which only one vaporization stage (or one exposed liquid surface) is involved.

Simple batch distillation is done by boiling a liquid mixture in a steam-jacketed-kettle (pot) and the vapor generated is withdrawn and condensed (distillate) as fast as it forms so that the vapor and the liquid do not have sufficient time to reach its equilibrium.

The first portion of vapor condensed will be richest in the more volatile component A. As the vaporization proceeds, the vaporized product becomes leaner in A.

SIMPLE BATCH DISTILLATION

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Raleigh equation for ideal and non – ideal mixtures

Consider a typical differential distillation at an instant time, t1 as shown below:

Where L = number of moles of liquid mixture in the pot x = mol fraction of a more – volatile component (component A) dx = infinitesimal change of mol fraction of A


Now consider that the differential distillation at certain infinitesimal time lapse (t) at t2 where t2 = t1 + t, after an infinitesimal amount of liquid has vaporized as shown below:

Mass balance of component A will yield:


dLydLLdxxxL

Negligible Cancel out

ydLdxdLLdxxdLxLxL

ydLxdLLdx

dLxyLdx

xy

dx

L

dL

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Integrating both sides within limits initial t1 and final t2:

Eq. (4) Where L1 = number of moles of liquid at t1

L2 = number of moles of liquid at t2

x1 = mol fraction of A in liquid at t1 x2 = mol fraction of A in liquid at t2

Eq. (4) known as Raleigh equation – applicable for both an ideal solution ( constant) or non – ideal solution ( non - constant).

The term must be evaluated graphically by determining an area under

the graph of versus x between the limit x1 and x2.


1x

2x

1L

2L

dxxy

1dL

L

1

1x

2x

1L

2Ldx

xy

1Lln

1x

2x

21 dxxy

1LL lnln

1x

2x2

1 dxxy

1

L

Lln

1x

2x2

1 dxxy

1

L

Lln

1x

2x2

1 dxxy

1

L

Lln

Simplified Raleigh equation for ideal mixtures

For an ideal solution ( constant), the Raleigh equation can be simplified.

Consider a simple batch distillation process at an initial time, t1 as shown below:

L1 = number of moles of binary liquid mixture containing A and B at t1

A1 = number of moles component A in L1 at t1

B1 = number of moles component B in L1 at t1

x1 = mol fraction of A in in L1 at t1

dL = infinitesimal amount of liquid that has vaporized dA = infinitesimal amount of A that has vaporized dB = infinitesimal amount of B that has vaporized


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We know from definitions,

Since AB is constant for an ideal mixture,

After simplifying,

Rearranging,


dBdA

dAyA

dBdA

dByB

BA

AxA

BA

BxB

BA

A

dBdA

dBBA

B

dBdA

dA

xy

xy

BB

AAAB

dBA

dABAB

A

dA

B

dBAB

Integrating within the limits of t1 and t2,

Since AB is constant,

Eq. (5)

Equation 5 known as simplified Raleigh equation for simple batch distillation which applicable for ideal solution.


2

1

2

1

A

A

B

B

ABA

dA

B

dB

2

1

2

1

A

A

B

B

ABA

dA

B

dB 2

1

2

1

A

A

B

BAB AB lnln

1

2

1

2AB

A

A

B

Blnln

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A mixture of 100 mol containing 50 mol% n-pentane and 50 mol% n-heptane is distilled under differential (batch) conditions at 101.3 kPa until 40 mol is distilled.

What is the average composition of the total vapor distilled and the composition

of the liquid left? The equilibrium data as follows, where x and y are mole fractions of n-pentane:


Example 1

xA yA

1.000 1.000

0.867 0.984

0.594 0.925

0.398 0.836

0.254 0.701

0.145 0.521

0.059 0.271

0 0


Solution

on.distillatibatch of

end at the L liquid theofn compositio theis xunknown, The

1510.0

1

60

100ln

1ln

,4 Eq. into ngSubstituti

mol 60L 40-100V-LL VLL

:balance material From

mol 40distilled) V(mol mol 100LGiven

22

5.0

2

5.0

2

1

22

1

21221

1

x

x

x

x

dxxy

dxxy

dxxyL

L

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Solution

av

av

2av

2211avav2211

2

2

5.0

2

y is distilled vapor totalofn compositio Average

?y

40

60x1000.5y

V

LxLxy VyLxLx

component, volatilemoreon balance material fromThen

liquid. theofn compositio thebe willx

. xfind 0.510curveunder area graph, theFrom

curve. under the area of value theto

referring is value1

the x, versus1

graph plottingBy

value

dxxyxy

x

Fractionating Distillation

Fractionation – or stage distillation with reflux, from a simplified point of view, can be considered to be a process in which a series of flash – vaporization stages are arranged in a series in such manner that the vapor and liquid products from each stage flow counter – current to each other.

In contrast with flash dist., continuous dist. process is more suitable for mixtures of about the same volatility and the condensed vapor and residual liquid are more pure (since it is re-distilled).

The fractionator consists of many trays which have holes to permit the vapor, V which rises up from the lower tray to bubble through and mixes with the liquid, L on the upper tray and equilibrated, and V and L stream leave in equilibrium.

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Continuous / Fractionating Distillation

During the mixing, the vapor will pick up more of component A from the liquid while the liquid will be richer with component B. As the vapor rises further up, it becomes richer and richer in component A but poorer with component B.

Conversely, as the liquid falls further down, it becomes poorer with A but richer in B. Thus we obtain a bottom product and an overhead product of higher purity in comparison to those obtained by single – stage simple batch or flash distillation.

Note: fractionation refers to a process where a part or whole of distillate is being recycled back to the fractionator. The recycled distillation (known as reflux) will supply the bulk of liquid needed to mix with vapor.


Where; F = Feed flow rate (kmols/hr) xf = m.v.c. composition of feed (mole fraction or mol percentage) V = Vapour flow rate (kmols/hr) L = Reflux flow rate (kmols/hr) D = Top Product flow rate (kmols/hr) xD = m.v.c. composition of top vapour stream, top product, and reflux (mole fraction or mol percentage) V" = Reboiler exit stream flow rate (kmols/hr) W = Bottom-product flow rate (kmols/hr) xW = m.v.c. composition of bottom product and feed to reboiler (mole fraction or mol percentage)

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Consider of fractionating column of N plates, where the condenser and reboiler are counted as ‘plates’. A typical ‘nth’ plate has the streams shown below associated with:


The feed stream is introduced on some intermediate tray where the liquid has approximately the same composition as the feed. The part of column above feed point – rectifying section; the lower portion – stripping section.

With constant molal overflow assumption: Ln-1 = Ln = Ln+1 = ... etc. Vn-1 = Vn = Vn+1 = ... etc. Conditions for Constant Molal Overflow

Heat losses negligible (achieved more easily in industrial columns)

Negligible heat of mixing

Equal or close heats of vaporisation

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Number of Plates Required in a Distillation Column

Four streams are involved in the transfer of heat and material across a plate, as shown in figure above: Plate n receives liquid Ln+1 from plate n+1 above, and vapor Vn-1 from plate n-1 below. Plate n supplies liquid Ln to plate n-1, and vapor Vn to plate n+1.

Action of the plate is to bring about mixing so that the vapor Vn, of composition yn, reaches equilibrium with the liquid Ln, of composition xn.

NUMBER OF PLATES BY LEWIS – SOREL METHOD

NUMBER OF PLATES BY Mc – CABE & THIELE


CALCULATION FOR NUMBER OF PLATES

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From figure above, binary feed F is distilled to give a top product D and a bottom product W, with xf , xd , and xw as the corresponding mole fraction of more volatile component.

The vapor Vt rising from top plate is condensed, and part is run back as liquid at its boiling point to the column as reflux, the remain withdrawn as product.


A material balance above plate n gives:

Since moles of liquid overflow are constant, Ln may be replaced by Ln+1

and:

d

n

n

n

nn

dnnnnnn

xV

Dx

V

Ly

DxxLVyDLV

11

111 Balance) (M.V.C

6 Eq. d

n

n

n

nn x

V

Dx

V

Ly 1

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A material balance for bottom to above plate m (indicated by loop II) and note that Lm = Lm+1.

Eq. 6 (above feed point) and Eq. 7 (below feed point) – operating lines.

Equilibrium data are used to find the composition of the vapor above the liquid, and the enrichment line to calculate the composition of liquid on the next plate.

7 Eq.

Balance)(M.V.C

w

m

m

m

mm

wmmmmmm

xV

Wx

V

Ly

WxxLVyWVL

1

1


Example

A mixture of benzene and toluene containing 40 mole% benzene is to be separated to give a product of 90 mole% of benzene at the top, and a bottom product with not more than 10 mole% of benzene. The feed is heated so that it enters the column at its boiling point, and the vapor leaving the column is condensed but not cooled, and provides reflux and product.

It is proposed to operate the unit with a reflux ratio of 3 kmol/kmol product. It is required to find the number of theoretical plates needed and the position of entry for the feed. The equilibrium diagram for operating at 1 bar pressure is shown in Figure below.

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Solution Feed : 40 mol% benzene (A), xf = 0.40 Product : 90 mol% benzene (A), xd = 0.90 Bottom : 10 mol% benzene (A), xw = 0.10

Do a basis: 100 kmol of feed. A total material balance gives: A balance on M.V.C (benzene) gives:

a Eq. D-100W WD100 WDF

kmol 62.5W

37.5-100 W,a Eq. From

kmol. 37.5D

D-1000.10.9D40

b Eq. into a Eq. Sub

b Eq. 0.1W 0.9D40

0.1W0.9D0.4100 xWxDxF wdf


Using a notation from reflux,

0.225 0.75

0.9150

37.5

150

112.5

:6) (Eq.equation line operating top theThus,

kmol 150V 37.5112.5V DLV

stage, at top balance Material From

5.112L 37.53L 3DL RDL D

LR

1

11

nnnn

nnnnn

nn

nnd

n

n

n

nn

xy

xyxV

Dx

V

Ly

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Since the feed is all liquid at its boiling point, it will all run down as increased reflux to the plate below. Thus:

0.04171.417

0.1150

62.5

150

212.5

line, operating BottomFrom

Vkmol 150V 62.5212.5V WLV

Bottom,at balancematerial From

kmol 212.5L 100112.5L FLL

mmmm

mmnm

1

11

mm

mmw

m

m

m

mm

n

xy

xyxV

Wx

V

Ly


Since all vapor from column is condensed, the composition of the vapor, yt from top plate = product, xd and liquid returned, xr

The composition xt of the liquid on the top plate is found from the equilibrium curve and since it is in equilibrium with vapor of composition yt = 0.90, xt = 0.79.

0.9 0.2250.9 0.75

0.9 xx say, Let

0.225 0.75 From

d1n

nn

nn

yy

xy 1

0.644x 0.8175, at diagram Eq. From

0.8175

0.225 0.790.75 0.225 0.75

1t

1

1

11

t

t

ttt

y

y

yxy

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0.594

0.225 0.4920.75 0.225 0.75


0.708

0.225 0.6440.75 0.225 0.75

3t

2t

3

3

323

2

2

212

t

t

ttt

t

t

ttt

y

y

yxy

y

y

yxy

This last value of composition is sufficiently near to the feed composition (xf =0.4). Feed to be introduced on plate t-3.

For the lower part of the column, the bottom operating line equation will be used.


0.048 x0.127 yat diagram, Eq. From

0.127 0.0417 0.1201.417

0.04171.417

0.120 x 0.252 yat diagram, Eq. From

0.252 0.0417 0.2081.417

0.04171.417

0.208 x 0.379 yat diagram, Eq. From

0.379 0.0417 0.2981.417

0.04171.417

0.298 x 0.50, yat diagram, Eq. From

0.50 0.0417 1.417

0.04171.417

0.04171.417

7-t7t

6-t6t

5-t5t

4-t4t

77

67

66

56

55

45

44

34

1

3820

tt

tt

tt

tt

tt

tt

tt

tt

mm

yy

xy

yy

xy

yy

xy

y.y

xy

xy

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This liquid xt-7 is slightly weaker than the minimum required and maybe withdrawn as the bottom product (the xt-7 value < xw).

Thus, xt-7 will correspond to the reboiler, and there will be seven plates in the column.

No of theoretical stages = 7 stage + 1 reboiler.


yt = 0.9

xt = 0.79

yt-1 = 0.8175

xt-1 = 0.644

yt-2 = 0.708

xt-2 = 0.492

yt-3 = 0.594

xt-3 = 0.382

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Assumption: constant molal heat of vaporization, no heat losses, no heat on mixing, led to a constant molal vapor flow and constant molar reflux flow in any section of column, the two enrichment equation obtained were:

Mc Cabe and Thiele pointed out that, since these equations represent straight lines connecting yn with xn+1 and ym with xm+1, they can be drawn on the same diagram as the equilibrium curve to give a simple graphical solution for the number of stages required .

In Eq. (6), if xn+1 = xd, then yn = xn+1 = xd :


6 Eq. d

n

n

n

nn x

V

Dx

V

Ly 1

7Eq. w

m

m

m

mm x

V

Wx

V

Ly 1

d

n

nd

n

nd

n

ndd

n

d

n

nn x

V

Vx

V

DLx

V

DLxx

V

Dx

V

Ly

1

In Eq. (6), if xn+1 = 0, then yn = Dxd / Vn :

The top operating line drawn through two points of coordinates (xd, xd)

and (0, Dxd / Vn) In Eq. (7), if xm+1 = xw, then ym = xw.

For bottom operating line passes point (xw, xw) and has a slope (Lm/Vm). When two operating lines drawn in, the number of stages required may

be found by drawing steps between the operating line and the equilibrium curve starting from point xd.

Limitation of method: 1.3 < relative volatility < 5, reflux ratio >1.1 minimum

reflux ratio, theoretical stages < 25


d

n

d

nn

nn x

V

Dx

V

D

V

Ly 0

ww

m

mw

m

mw

m

w

m

mm xx

V

Vx

V

WLx

V

Wx

V

Ly

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Example

A mixture of benzene and toluene containing 40 mole% benzene is to be separated to give a product of 90 mole% of benzene at the top, and a bottom product with not more than 10 mole% of benzene. The feed is heated so that it enters the column at its boiling point, and the vapor leaving the column is condensed but not cooled, and provides reflux and product.

It is proposed to operate the unit with a reflux ratio of 3 kmol/kmol product. It is required to find the number of theoretical plates needed and the position of entry for the feed. The equilibrium diagram for operating at 1 bar pressure is shown in Figure below.


Solution Feed : 40 mol% benzene (A), xf = 0.40 Product : 90 mol% benzene (A), xd = 0.90 Bottom : 10 mol% benzene (A), xw = 0.10

Do a basis: 100 kmol of feed. A total material balance gives: A balance on M.V.C (benzene) gives:

a Eq. D-100W WD100 WDF

kmol 62.5W

37.5-100W a Eq. From

kmol. 37.5D

D-1000.10.9D40

b Eq. into a Eq. Insert

b Eq. 0.1W0.9D40

0.1W0.9D0.4100 xWxDxF xdf

,


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Using a notation from reflux,

0.225 0.75

0.9150

37.5

150

112.5

:6) (Eq.equation line operating top theThus,

kmol 150V 37.5112.5V DLV

stage, at top balance Material From

5.112L 37.53L 3DL RDL D

LR

1

11

nnnn

nnnnn

nn

nnd

n

n

n

nn

xy

xyxV

Dx

V

Ly


Since the feed is all liquid at its boiling point, it will all run down as increased reflux to the plate below. Thus:

0.04171.417

0.1150

62.5

150

212.5

line, operating BottomFrom

Vkmol 150V 62.5212.5V WLV

Bottom,at balancematerial From

kmol 212.5L 100112.5L FLL

mmmm

mmnm

1

11

mm

mmw

m

m

m

mm

n

xy

xyxV

Wx

V

Ly


𝑴𝒂𝒔𝒔 𝑩𝒂𝒍𝒂𝒏𝒄𝒆 𝒐𝒗𝒆𝒓 𝒕𝒉𝒆 𝒇𝒆𝒆𝒅𝒊𝒏𝒈 𝒑𝒍𝒂𝒕𝒆:- 𝐹 + 𝐿𝑛 + 𝑉𝑚 = 𝐿𝑚 + 𝑉𝑛 or 𝑉𝑚 − 𝑉𝑛 = 𝐿𝑚-𝐿𝑛-𝐹 𝐸𝑞 (9)

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Feed is form of liquid at its boiling point, the reflux overflowing . If feed is a liquid at temperature Tf, less than the boiling point, some vapor rising from the plate below will condense to provide sufficient heat to bring the feed liquor to the boiling point.

Let Hf = enthalpy per mol of feed, Hfs = enthalpy of one mol of feed at its boiling point. Heat to be supplied to bring feed to its boiling point is F(Hfs - Hf ), number of vapor mols to be condensed to provide this heat is F(Hfs - Hf ) / , where is molar latent heat of vapor

Reflux liquor:


FLn

feed of heat latent Molar

feed of mol 1 vaporize to Heat

10Eq.

q

qFLLHHF

LLHHF

FLL nm

ffs

nm

ffs

nm

Insert Eq 9 into 10:

A material balance for m.v.c over whole column gives:

From Eq. 8, Eq. 10 and Eq 11:

If xq = xf, then yq = xf. The point of intersection of the two operating lines lies on the

straight line of slope q / (q-1) passing through point (xf, xf). When yq = 0, xq = xf / q. The line may thus be drawn through two easily determined points.


11Eq. FqFVVFLqFLVV nmnnnm

wdf WxDxFx

equation line-q 12Eq.

11

1

q

xx

q

qy

FxxqFFqyFxxqFFqFy

f

qq

fqqfqq

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From definition of q, it follows that the slope of q – line is governed by the nature of feed,


Total and Minimum Reflux Ratio for Mc Cabe – Thiele Method

Total reflux –Limiting values of the reflux ratio, R = . Since R = Ln/D and Vn=Ln+1+D, Ln+1 is very large, as is the vapor Vn. The slope of the top operating line, = 1.0, = 0, eq. become, operating line of both sections of the column coincide with the 45O diagonal line as shown in figure below .


section) g(rectifyin line operating Top

11

1R

xx

R

Ry d

nn

1R

R

1R

xd

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The number of theoretical trays required obtained as before by stepping off the trays from the distillate to the bottoms – gives the minimum number of trays can possibly be used to obtain the given separation.

In actual practice – condition can be realized by returning all the overhead condenser vapor V1 from the top of the tower back to the tower as reflux, that is total reflux. All the liquid in the bottom is reboiled – all the production distillate and bottoms are reduced to zero flow, as the fresh feed to the tower.


Minimum reflux ratio – reflux ratio, Rm that will acquire an infinite number of trays for the given desired separation of xd and xw. This corresponds to the minimum vapor flow in the tower, hence minimum reboiler and condenser size.

If R decreased, the slope of AE (R/(R+1)) is decreased, AE closer to the equilibrium line. Number of steps required to give a fixed xd and xw increases.

When two operating line touch the equilibrium line, a “pinch point” at y’ and x’ occurs where the number of steps required become infinite.

The slope of AE:

The y-intercept of AE:


'xx

'yx

R

R

d

d

m

m

1

1

m

d

R

xy intercept

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31

Operating and optimum reflux ratio:

For the case of total reflux, the number of plates is a minimum, but the

tower diameter is infinite. This corresponds to an infinite cost of tower.

For minimum reflux, the number of trays infinite, gives an infinite cost.

These are the two limits in operation of the tower. The actual reflux

ratio to use is lies between two limits. The optimum reflux ratio between Rm total reflux. For many cases,

operating reflux ratio between 1.2Rm and 1.5Rm.


Example 2

11,000 kg/h of equal parts (in wt) of Benzene – Toluene solution is to be distilled in a fractionating tower at atmospheric pressure.

The liquid is fed as a liquid – vapor mixture in which the feed consist of 75% vapor. The distillate contains 94 wt% Benzene whereas the bottom products contains 98 wt% toluene. Using the equilibrium diagram of Benzene-Toluene mixture, determine by McCabe-Thiele’s method:

The flow rate of distillate and bottom product in kg/h The minimum reflux ratio, Rm. The number of theoretical stages required if the reflux ratio

used is 2 times the minimum reflux ratio The position of the feed tray.

Data: Molecular weight of Benzene = 78 Molecular weight of Toluene = 92


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Solution 2 Feed : 50 wt% benzene (A), xf = 0.50 F : 11,000 kg/h Product : 94 wt% benzene (A), xd = 0.94 Bottom : 2 wt% benzene (A), xw = 0.02

The flow rate of distillate and bottom product in kg/h

A total material balance gives: A balance on M.V.C (benzene) gives:

a Eq. D-11000W WD11000 WDF

kg/h 5260.9W

5739.1-11000W a Eq. From

kg/h. 5739.1D

D-110000.020.94D5500

b Eq. into a Eq. Insert

b Eq. 0.02W0.94D5500

0.02W0.94D0.511000 xWxDxF wdf

,


The minimum reflux ratio, Rm.


1) Convert mass fraction to mol fraction Do basis calculation: 100 kg

Feed (F) Distillate (D) Bottoms (W)

Component Mwt Mass Mol Mol frac.

Mass Mol Mol frac.

Mass Mol Mol frac.

Benzene (A) 78 50 0.64 0.54 94 1.21 0.95 2 0.03 0.03

Toluene (B) 92 50 0.54 0.46 6 0.07 0.05 98 1.07 0.97

Total 1.18 1.00 1.28 1.00 1.10 1.00

From conversion table, mol fraction for M.V.C: xf = 0.54 xd = 0.95 xw = 0.03

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2) Find q – line. Feed enters at 75% vapor.

0.62 0.720.30.333 0.3, x Let

0.720.333 10.25

0.54

10.25

0.25

11

equation line

0.25 (1.0) 0.25 (0) 0.75

)( 0.25 )( 0.75 liq.vap.

qq

qqqq

f

qq

yy

xyxy

q

xx

q

qy

q

qq

qqq

Plot (0.54 , 0.54) and (0.30 , 0.62) for q-line


From the graph, y intercept for q – line = 0.36

1.64 0.361

0.95 0.36

m

mm

d RRR

x

1

The number of theoretical stages required if the reflux ratio used

is 2 times the minimum reflux ratio

0.605 0.2220.50.766 0.5, x At

0.2220.766

13.28

0.95

13.28

3.28

line operating Top 1

3.28 1.642 2 min

nn

nn

nn

dnn

yy

xy

xy

R

xx

R

Ry

RRR

1

1

11

Plot (0.95 , 0.95) and (0.5, 0.605) for T.O.L

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The number of theoretical stages required = 10.5 stages including

reboiler

Feed plate location: 5 from top

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DISTILLATION - جامعة نزوى · 24/02/2015 3 VAPOR – LIQUID ... Flash distillation is done by vaporizing a definite fraction of a liquid mixture in such a way that the evolved